I came upon a part of a proof in the script of my professor about commutative algebra where explicit explanation is left out. The theorem is the following:
Let $R$ be a ring (commutative with identity), let $M^{\prime}$, $M$ and $M^{\prime\prime}$ be $R$-modules, and let $f:M^{\prime} \to M$ and $g:M\to M^{\prime\prime}$ be $R$-module homomorphisms. Then the following statements are equivalent:
(i) The sequence $M^{\prime} \to M\to M^{\prime\prime}\to 0$ is right exact.
(ii) For every $R$-module $N$ the induced $R$-module homomorphisms $Hom(M,N) \to Hom(M^{\prime},N)$ and $Hom(M^{\prime\prime},N) \to Hom(M,N)$ fit into a left exact sequence $$0\to Hom(M^{\prime\prime},N) \to Hom(M,N) \to Hom(M^{\prime},N)$$
Now in the proof for (i) $\implies$ (ii), there is written:
"Since $g\circ f = 0$, the composite $Hom(M^{\prime\prime},N) \to Hom(M^{\prime},N)$ is zero."
So let us take $\theta \in Hom(M^{\prime\prime},N)$ and $\phi \in Hom(M^{\prime},N)$, i.e. $\theta: M^{\prime\prime} \to N$ and $\phi: M^{\prime} \to N$. Why can't I send $\theta$ to some nonzero $\phi$? In other words, I'm irritated by the formulation "the composite (...) is zero"; aren't there other ways to fit the induced homomorphisms into sequences that are not left-exact, but the theorem states that there also exists one embedding such that the sequence becomes left exact?
Presumably $f$ is the map $M' \to M$ and $g$ is the map $M \to M''$. By the assumption of right exactness, $g \circ f = 0$.
So what does the composite $Hom(M'', N) \to Hom(M' N)$ look like? The induced map $Hom(M'', N) \to Hom(M, N)$ sends $\theta: M'' \to N$ to $\theta \circ g: M \to N$, and the induced map $Hom(M, N) \to Hom(M', N)$ sends $\psi: M \to N$ to $\psi \circ f: M' \to N$. So the composite $Hom(M'', N) \to Hom(M', N)$ sends $\theta$ to $\theta \circ g \circ f = \theta \circ 0 = 0$.